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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 17: Temperature and HeatProblem 38
Chapter 17, Problem 38

A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

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Identify the materials involved: copper calorimeter, water, ice, and lead. Note their masses and initial temperatures.
Use the principle of conservation of energy, which states that the total heat lost by the lead will be equal to the total heat gained by the copper calorimeter, water, and ice.
Calculate the heat required to melt the ice using the formula: Q=mL, where m is the mass of the ice and L is the latent heat of fusion for ice.
Calculate the heat exchange for each material using the formula: Q=mc(Tf-Ti), where m is the mass, c is the specific heat capacity, Tf is the final temperature, and Ti is the initial temperature.
Set up the equation for conservation of energy: the sum of heat gained by the copper, water, and ice equals the heat lost by the lead. Solve for the final temperature Tf.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. It is crucial in this problem to calculate how much heat is absorbed or released by each material (copper, water, ice, and lead) as they reach thermal equilibrium. Different materials have different specific heat capacities, affecting how they exchange heat.
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Heat Transfer and Thermal Equilibrium

Heat transfer is the process of energy moving from a hotter object to a cooler one until thermal equilibrium is reached, meaning all objects involved are at the same temperature. In this scenario, the lead, initially at a higher temperature, will transfer heat to the copper, water, and ice until they all reach a common final temperature. Understanding this concept helps in setting up the energy balance equation.
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Phase Change and Latent Heat

Phase change involves a substance transitioning between solid, liquid, or gas phases, requiring or releasing latent heat without changing temperature. In this problem, the ice may melt, requiring latent heat of fusion. This concept is essential to account for the energy needed to change the ice to water before it can further increase in temperature, affecting the final temperature calculation.
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Related Practice
Textbook Question

One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at 0°C to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached 32.0°C. To treat a 70.0 kg patient, what is the minimum amount of ice (at 0°C) you need in the bath so that its temperature remains at 0°C? The specific heat of the human body is 3480 J/kg C°, and recall that normal body temperature is 37.0°C.

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Textbook Question

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0°C. A 0.250-kg block of iron at 85.0°C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Textbook Question

A blacksmith cools a 1.20 kg chunk of iron, initially at 650.0°C, by trickling 15.0°C water over it. All of the water boils away, and the iron ends up at 120.0°C. How much water did the blacksmith trickle over the iron?

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Textbook Question

Before going in for his annual physical, a 70.0 kg man whose body temperature is 37.0°C consumes an entire 0.355-L can of a soft drink (mostly water) at 12.0°C. What will his body temperature be after equilibrium is attained? Ignore any heating by the man’s metabolism. The specific heat of the man’s body is 3480 J/kg K.

Textbook Question

An ice-cube tray of negligible mass contains 0.290 kg of water at 18.0°C. How much heat must be removed to cool the water to 0.00°C and freeze it? Express your answer in joules, calories, and Btu.

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Textbook Question

You have 750 g of water at 10.0°C in a large insulated beaker. How much boiling water at 100.0°C must you add to this beaker so that the final temperature of the mixture will be 75°C?

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